C Program - Find all factors of a Number

Objective: Write a C program to find all distinct factors (divisors) of a given natural number. The divisors of few numbers are given below:

Number: 10
Divisors: 1 2 5 10

Number: 15
Divisors: 1 3 5 15

Number: 100
Divisors: 1 2 4 5 10 20 25 50 100

Method 1: Using iteration

One of the basic approach is to iterate from 1 to n and in each iteration check whether the number divides n. If it divides then print it.

#include <stdio.h>

//function to print all divisors of a number
static void printDivisors(int n) {
printf("Divisors of %i are: ", n);
for(int i = 1; i <= n; i++) {
if(n%i == 0)
printf("%i ", i);
}
printf("\n");
}

int main(){
printDivisors(10);
printDivisors(50);
printDivisors(100);
return 0;
}

The above code will give the following output:

Divisors of 10 are: 1 2 5 10
Divisors of 50 are: 1 2 5 10 25 50
Divisors of 100 are: 1 2 4 5 10 20 25 50 100

Method 2: Optimized Code

Instead of checking the divisibility of the given number from 1 to n, it is checked till square root of n. For a factor larger than square root of n, there must the a smaller factor which is already checked in the range of 1 to square root of n.

#include <stdio.h>
#include <math.h>

//function to print all divisors of a number
static void printDivisors(int n) {
printf("Divisors of %i are: ", n);
//loop from 1 to sqrt(n)
for(int i = 1; i <= sqrt(n); i++) {
if(n%i == 0) {
if(n/i == i)
printf("%i ", i);
else
printf("%i %i ", i, n/i);
}
}
printf("\n");
}

int main(){
printDivisors(10);
printDivisors(50);
printDivisors(100);
return 0;
}

The above code will give the following output:

Divisors of 10 are: 1 10 2 5
Divisors of 50 are: 1 50 2 25 5 10
Divisors of 100 are: 1 100 2 50 4 25 5 20 10

Method 3: Optimized Code with sorted result

In the previous method, results are produced in a irregular fashion (printed in pairs - small number and large number). The result can be sorted by storing the larger number and print them later on. Consider the example below:

#include <stdio.h>
#include <math.h>

//function to print all divisors of a number
static void printDivisors(int n) {
printf("Divisors of %i are: ", n);

//creating an array to store larger numbers
int arr[n];
int j = 0;

//loop from 1 to sqrt(n)
for(int i = 1; i <= sqrt(n); i++) {
if(n%i == 0) {
if(n/i == i)
printf("%i ", i);
else {
printf("%i ", i);
//storing the large number of a pair
arr[j++] = (n/i);
}
}
}

//printing stored large numbers of pairs
for(int i = j - 1; i >= 0; i--)
printf("%i ", arr[i]);
printf("\n");
}

int main(){
printDivisors(10);
printDivisors(50);
printDivisors(100);
return 0;
}

The above code will give the following output:

Divisors of 10 are: 1 2 5 10
Divisors of 50 are: 1 2 5 10 25 50
Divisors of 100 are: 1 2 4 5 10 20 25 50 100

Method 4: Another Optimized Code

To produce the result in sorted order, we can iterate from 1 to square root of n and printing the number which divides n. After that we can iterate back (in reverse order) and printing the quotient of all numbers which divides n.

#include <stdio.h>
#include <math.h>

//function to print all divisors of a number
static void printDivisors(int n) {
printf("Divisors of %i are: ", n);
//loop from 1 to sqrt(n)
int i;
for(i = 1; i <= sqrt(n); i++) {
if(n%i == 0)
printf("%i ", i);

//handing perfect squares
if(n/i == i) {
i--; break;
}
}

for(; i >= 1; i--) {
if(n%i == 0)
printf("%i ", n/i);
}
printf("\n");
}

int main(){
printDivisors(10);
printDivisors(50);
printDivisors(100);
return 0;
}

The above code will give the following output:

Divisors of 10 are: 1 2 5 10
Divisors of 50 are: 1 2 5 10 25 50
Divisors of 100 are: 1 2 4 5 10 20 25 50 100

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